Chapter-3: Elementary Functions

Dr. Jajati Keshari Sahoo

Department of Mathematics
BITS Pilani K.K. Birla Goa Campus

June 06, 2023

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Examples

Example. Find the

(a) complex exponents i i and principal value of i i .
(b) complex exponents (1 − i)4i .

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Examples

Example. Find the

(a) complex exponents i i and principal value of i i .
(b) complex exponents (1 − i)4i .
Solution:(A)

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Examples

Example. Find the

(a) complex exponents i i and principal value of i i .
(b) complex exponents (1 − i)4i .
Solution:(A)
Since i i = e i log i ,and log i = ln |i| + i π2 + 2nπi = ( 21 + 2n)πi.

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Examples

Example. Find the

(a) complex exponents i i and principal value of i i .
(b) complex exponents (1 − i)4i .
Solution:(A)
Since i i = e i log i ,and log i = ln |i| + i π2 + 2nπi = ( 21 + 2n)πi.
1 1
So the complex exponent is e i×( 2 +2n)πi = e −( 2 +2n)π .

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Examples

Example. Find the

(a) complex exponents i i and principal value of i i .
(b) complex exponents (1 − i)4i .
Solution:(A)
Since i i = e i log i ,and log i = ln |i| + i π2 + 2nπi = ( 21 + 2n)πi.
1 1
So the complex exponent is e i×( 2 +2n)πi = e −( 2 +2n)π .
For the principal part we have to take Logi = π2 i and hence the principal
part of the exponent

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Examples

Example. Find the

(a) complex exponents i i and principal value of i i .
(b) complex exponents (1 − i)4i .
Solution:(A)
Since i i = e i log i ,and log i = ln |i| + i π2 + 2nπi = ( 21 + 2n)πi.
1 1
So the complex exponent is e i×( 2 +2n)πi = e −( 2 +2n)π .
For the principal part we have to take Logi = π2 i and hence the principal
π π
part of the exponent i i = e i 2 i = e − 2
Solution:(B)

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Examples

Example. Find the

(a) complex exponents i i and principal value of i i .
(b) complex exponents (1 − i)4i .
Solution:(A)
Since i i = e i log i ,and log i = ln |i| + i π2 + 2nπi = ( 21 + 2n)πi.
1 1
So the complex exponent is e i×( 2 +2n)πi = e −( 2 +2n)π .
For the principal part we have to take Logi = π2 i and hence the principal
π π
part of the exponent i i = e i 2 i = e − 2
Solution:(B) √
Since (1 − i)4i = e 4i log(1−i) and log(1 − i) = ln 2 + (− π4 + 2nπ)i

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Examples

Example. Find the

(a) complex exponents i i and principal value of i i .
(b) complex exponents (1 − i)4i .
Solution:(A)
Since i i = e i log i ,and log i = ln |i| + i π2 + 2nπi = ( 21 + 2n)πi.
1 1
So the complex exponent is e i×( 2 +2n)πi = e −( 2 +2n)π .
For the principal part we have to take Logi = π2 i and hence the principal
π π
part of the exponent i i = e i 2 i = e − 2
Solution:(B) √
Since (1 − i)4i = e 4i log(1−i) and log(1 − i) = ln 2 + (− π4 + 2nπ)i
√
2+(− π4 +2nπ)i
So the exponent is e 4i(ln and the principal part is e π+2 ln 2i .

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Branches of logarithm function

The multi-valued logarithm function is

log z = ln |z| + iθ, where θ = Arg(z) + 2πn, n ∈ Z. (1)

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Branches of logarithm function

The multi-valued logarithm function is

log z = ln |z| + iθ, where θ = Arg(z) + 2πn, n ∈ Z. (1)

For any α ∈ R, if we restrict θ in the interval α < θ < α + 2π, then, we

obtain asingle-valuedd logarithm function called branch and denoted by

log z = ln r + iθ, where r = |z|, r > 0, α < θ < α + 2π. (2)

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Branches of logarithm function

The multi-valued logarithm function is

log z = ln |z| + iθ, where θ = Arg(z) + 2πn, n ∈ Z. (1)

For any α ∈ R, if we restrict θ in the interval α < θ < α + 2π, then, we

obtain asingle-valuedd logarithm function called branch and denoted by

log z = ln r + iθ, where r = |z|, r > 0, α < θ < α + 2π. (2)

Principal Branch: If α = −π, then the branch is called principal branch

denoted by Log z. That is

Log z = ln r + iθ, where r = |z|, r > 0, −π < θ < π. (3)

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Branches of logarithm function

The multi-valued logarithm function is

log z = ln |z| + iθ, where θ = Arg(z) + 2πn, n ∈ Z. (1)

For any α ∈ R, if we restrict θ in the interval α < θ < α + 2π, then, we

obtain asingle-valuedd logarithm function called branch and denoted by

log z = ln r + iθ, where r = |z|, r > 0, α < θ < α + 2π. (2)

Principal Branch: If α = −π, then the branch is called principal branch

denoted by Log z. That is

Log z = ln r + iθ, where r = |z|, r > 0, −π < θ < π. (3)

Note that for principal both sides strictly inequality

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Analyticity of logarithm functions
Consider the single-valued logarithm function

log z = ln r + iθ, where r = |z|, r > 0, α < θ < α + 2π.

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Analyticity of logarithm functions
Consider the single-valued logarithm function

log z = ln r + iθ, where r = |z|, r > 0, α < θ < α + 2π.

In the polar coordinates, we have u(r , θ) = ln r and v (r , θ) = θ.

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Analyticity of logarithm functions
Consider the single-valued logarithm function

log z = ln r + iθ, where r = |z|, r > 0, α < θ < α + 2π.

In the polar coordinates, we have u(r , θ) = ln r and v (r , θ) = θ.

Since ur = 1/r , uθ = 0, vr = 0, vθ = 1. So it satisfies the CR equations in
polar form ( recall rur = vθ , uθ = −rvr .) Thus the derivative is :

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Analyticity of logarithm functions
Consider the single-valued logarithm function

log z = ln r + iθ, where r = |z|, r > 0, α < θ < α + 2π.

In the polar coordinates, we have u(r , θ) = ln r and v (r , θ) = θ.

Since ur = 1/r , uθ = 0, vr = 0, vθ = 1. So it satisfies the CR equations in
polar form ( recall rur = vθ , uθ = −rvr .) Thus the derivative is :

d 1 1
log z = e −iθ (ur + ivr ) = e −iθ (1/r + i0) = iθ = ,
dz re z
where |z| > 0, α < argz < α + 2π.
Similarly
d 1 1
Log z = e −iθ (ur + ivr ) = e −iθ (1/r + i0) = iθ = ,
dz re z
where |z| > 0, −π < argz < π.
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Nature of branches at the ray θ = α
Consider the principal branch at the ray θ = π ( ⇐⇒ z ∈ (−∞, 0)).
Log z = ln r + iθ, (−π < θ < π).

Figure: Nature of Log z on (−∞, 0)

Ray: Θ=π

-π

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Nature of branches at the ray θ = α

From the previous Figure, it is clearly true that the Arg z is not
continuous on (−∞, 0). Therefore, the principal branch Log z is not
continuous on (−∞, 0) or θ = π.

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Nature of branches at the ray θ = α

From the previous Figure, it is clearly true that the Arg z is not
continuous on (−∞, 0). Therefore, the principal branch Log z is not
continuous on (−∞, 0) or θ = π.

Similarly, other branches are not continuous at the respective ray

θ = α. Hence the branches are not analytic on the ray θ = α.

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Nature of branches at the ray θ = α

From the previous Figure, it is clearly true that the Arg z is not
continuous on (−∞, 0). Therefore, the principal branch Log z is not
continuous on (−∞, 0) or θ = π.

Similarly, other branches are not continuous at the respective ray

θ = α. Hence the branches are not analytic on the ray θ = α.

Also note that, the points lies on the ray θ = α are singular points of
the respective branches.

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Branch cut

Definition
A branch cut is a minimal set of points so that the branch considered can
be consistently defined by analytic continuation on the complement of the
branch cut.

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Branch cut

Definition
A branch cut is a minimal set of points so that the branch considered can
be consistently defined by analytic continuation on the complement of the
branch cut.

Note: Points on the branch cut are singular points of the respective
branch.

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Branch cut

Definition
A branch cut is a minimal set of points so that the branch considered can
be consistently defined by analytic continuation on the complement of the
branch cut.

Note: Points on the branch cut are singular points of the respective
branch.
Example
Consider the logarithm function,
log z = ln r + iθ, (α < θ < α + 2π, α ∈ R). Then each ray θ = α is the
branch cut of log z

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Branch cut

Definition
A branch cut is a minimal set of points so that the branch considered can
be consistently defined by analytic continuation on the complement of the
branch cut.

Note: Points on the branch cut are singular points of the respective
branch.
Example
Consider the logarithm function,
log z = ln r + iθ, (α < θ < α + 2π, α ∈ R). Then each ray θ = α is the
branch cut of log z

Note: θ = π or the interval (−∞, 0) is the branch cut of Log z.

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Branch Point

Definition
A point which is common to all branch cuts is called a branch point.

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Branch Point

Definition
A point which is common to all branch cuts is called a branch point.

Example
Consider the logarithm function,
log z = ln r + iθ, (α < θ < α + 2π, α ∈ R). Since each ray starting at 0,
is a branch cut, so z = 0 is the branch point of log z.

Example
Consider the logarithm function,
log (z − 1) = ln |(z − 1)| + iarg (z − 1), (α < arg (z − 1) < α + 2π, α ∈ R).
Since each ray starting at 1, is a branch cut, so z = 1 is the branch point
of log (z − 1).

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Some more Properties

Since log z is analytic on α < θ < α + 2π, so the following are analytic on
the same domain:
(a) z c , c ∈ C.
(b) c z , c ∈ C.
d c d z
Also we have dz z = cz c−1 and dz c = c z log c.

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Outline

1 Trigonometric functions

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Trigonometric functions

One can motivate the above definitions from Euler’s identity,

e ix = cos x + i sin x and e −ix = cos x − i sin x .

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Trigonometric functions

One can motivate the above definitions from Euler’s identity,

e ix = cos x + i sin x and e −ix = cos x − i sin x . So we can add and subtract
to get formulas for sin x and cos x in terms of exponentials involving
complex numbers.

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Trigonometric functions

One can motivate the above definitions from Euler’s identity,

e ix = cos x + i sin x and e −ix = cos x − i sin x . So we can add and subtract
to get formulas for sin x and cos x in terms of exponentials involving
complex numbers.
Now we will define the complex trigonometric functions as follows:

e iz − e −iz
sin z =
2i

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Trigonometric functions

One can motivate the above definitions from Euler’s identity,

e ix = cos x + i sin x and e −ix = cos x − i sin x . So we can add and subtract
to get formulas for sin x and cos x in terms of exponentials involving
complex numbers.
Now we will define the complex trigonometric functions as follows:

e iz − e −iz
sin z =
2i
and
e iz + e −iz
cos z =
2

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Properties of sine and cosine functions

d
1
dz sin z = cos z.
d
2
dz cos z = − sin z.

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Properties of sine and cosine functions

d
1
dz sin z = cos z.
d
2
dz cos z = − sin z.
3 sin(−z) = − sin z and cos(−z) = cos z.
4 e iz = cos z + i sin z.

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Properties of sine and cosine functions

d
1
dz sin z = cos z.
d
2
dz cos z = − sin z.
3 sin(−z) = − sin z and cos(−z) = cos z.
4 e iz = cos z + i sin z.
5 sin2 z + cos2 z = 1.
6 sin(z) and cos(z) both are periodic with period 2π.

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Properties of sine and cosine functions

d
1
dz sin z = cos z.
d
2
dz cos z = − sin z.
3 sin(−z) = − sin z and cos(−z) = cos z.
4 e iz = cos z + i sin z.
5 sin2 z + cos2 z = 1.
6 sin(z) and cos(z) both are periodic with period 2π.
7 sin(z1 + z2 ) = sin z1 cos z2 + cos z1 sin z2 .

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Properties of sine and cosine functions

d
1
dz sin z = cos z.
d
2
dz cos z = − sin z.
3 sin(−z) = − sin z and cos(−z) = cos z.
4 e iz = cos z + i sin z.
5 sin2 z + cos2 z = 1.
6 sin(z) and cos(z) both are periodic with period 2π.
7 sin(z1 + z2 ) = sin z1 cos z2 + cos z1 sin z2 .
8 cos(z1 + z2 ) = cos z1 cos z2 − sin z1 sin z2 .

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Properties of sine and cosine functions

d
1
dz sin z = cos z.
d
2
dz cos z = − sin z.
3 sin(−z) = − sin z and cos(−z) = cos z.
4 e iz = cos z + i sin z.
5 sin2 z + cos2 z = 1.
6 sin(z) and cos(z) both are periodic with period 2π.
7 sin(z1 + z2 ) = sin z1 cos z2 + cos z1 sin z2 .
8 cos(z1 + z2 ) = cos z1 cos z2 − sin z1 sin z2 .
9 sin(z + π2 ) = cos z.

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Properties of sine and cosine functions

d
1
dz sin z = cos z.
d
2
dz cos z = − sin z.
3 sin(−z) = − sin z and cos(−z) = cos z.
4 e iz = cos z + i sin z.
5 sin2 z + cos2 z = 1.
6 sin(z) and cos(z) both are periodic with period 2π.
7 sin(z1 + z2 ) = sin z1 cos z2 + cos z1 sin z2 .
8 cos(z1 + z2 ) = cos z1 cos z2 − sin z1 sin z2 .
9 sin(z + π2 ) = cos z.
10 sin(z − π2 ) = − cos z

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Further Properties and Definitions

1 sin z = 0 if and only if z = nπ, n ∈ Z.

2 cos z = 0 if and only if z = (2n + 1) π2 , n ∈ Z.

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Further Properties and Definitions

1 sin z = 0 if and only if z = nπ, n ∈ Z.

2 cos z = 0 if and only if z = (2n + 1) π2 , n ∈ Z.
sin z cos z
3 tan z = cos z and cot z = sin z .

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Further Properties and Definitions

1 sin z = 0 if and only if z = nπ, n ∈ Z.

2 cos z = 0 if and only if z = (2n + 1) π2 , n ∈ Z.
sin z cos z
3 tan z = cos z and cot z = sin z .
1 1
4 sec z = cos z and csc z = sin z .

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Further Properties and Definitions

1 sin z = 0 if and only if z = nπ, n ∈ Z.

2 cos z = 0 if and only if z = (2n + 1) π2 , n ∈ Z.
sin z cos z
3 tan z = cos z and cot z = sin z .
1 1
4 sec z = cos z and csc z = sin z .
d
5
dz tan z = sec2 z.
d
6
dz cot z = − csc2 z.

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Further Properties and Definitions

1 sin z = 0 if and only if z = nπ, n ∈ Z.

2 cos z = 0 if and only if z = (2n + 1) π2 , n ∈ Z.
sin z cos z
3 tan z = cos z and cot z = sin z .
1 1
4 sec z = cos z and csc z = sin z .
d
5
dz tan z = sec2 z.
d
6
dz cot z = − csc2 z.
d
7
dz sec z = sec z tan z.
d
8
dz cot z = − csc z cot z.
9 tan(z + π) = tan z

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Example

Example
Find all roots of f (z) = cos z − 2 or solve cos z = 2.

Let cos z = 2. Which implies e iz + e −iz = 4. Thus

e 2iz − 4e iz + 1 = 0.

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Example

Example
Find all roots of f (z) = cos z − 2 or solve cos z = 2.

Let cos z = 2. Which implies e√iz + e −iz = 4. Thus

2iz iz iz iz
√
e − 4e + √ 1 = 0. e = 2 ± 3. Consider e = 2 + 3. This implies
e iz = e ln(2+ 3) .

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Example

Example
Find all roots of f (z) = cos z − 2 or solve cos z = 2.

Let cos z = 2. Which implies e√ iz + e −iz = 4. Thus

2iz iz iz iz
√
e − 4e + √ 1 = 0. e = 2 ± 3. Consider e √= 2 + 3. This implies
e iz = e ln(2+ 3) . Therefore
√ iz = 2nπi + ln(2 + 3). Hence
z = 2nπ − i ln(2 + 3).

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Example

Example
Find all roots of f (z) = cos z − 2 or solve cos z = 2.

Let cos z = 2. Which implies e√ iz + e −iz = 4. Thus

2iz iz iz iz
√
e − 4e + √ 1 = 0. e = 2 ± 3. Consider e √= 2 + 3. This implies
e iz = e ln(2+ 3) . Therefore
√ iz = 2nπi + ln(2 + 3). Hence
z = 2nπ − i ln(2 + 3). Similarly, from other value, we get
√ 1
  √
z = 2nπ − i ln(2 − 3) = 2nπ + i ln √ = 2nπ + i ln(2 + 3).
2− 3
Finally, we have √
z = 2nπ ± i ln(2 + 3).

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Hyperbolic functions

x −x
Using the real hyperbolic functions namely sinh x = e −e 2 and
e x +e −x
cosh x = 2 one can write the following identities for z = x + iy :

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Hyperbolic functions

x −x
Using the real hyperbolic functions namely sinh x = e −e 2 and
e x +e −x
cosh x = 2 one can write the following identities for z = x + iy :

sin z = sin x cosh y + i cos x sinh y

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Hyperbolic functions

x −x
Using the real hyperbolic functions namely sinh x = e −e 2 and
e x +e −x
cosh x = 2 one can write the following identities for z = x + iy :

sin z = sin x cosh y + i cos x sinh y

and

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Hyperbolic functions

x −x
Using the real hyperbolic functions namely sinh x = e −e 2 and
e x +e −x
cosh x = 2 one can write the following identities for z = x + iy :

sin z = sin x cosh y + i cos x sinh y

and
cos z = cos x cos hy − i sin x sinh y

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Hyperbolic functions

x −x
Using the real hyperbolic functions namely sinh x = e −e 2 and
e x +e −x
cosh x = 2 one can write the following identities for z = x + iy :

sin z = sin x cosh y + i cos x sinh y

and
cos z = cos x cos hy − i sin x sinh y
These hold because sin(iy ) = i sinh y and cos(iy ) = cosh y .
Exercise: Solve sin z = cosh 4.

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Exercise

Ex. Show that both sin z and cos z are unbounded.

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Exercise

Ex. Show that both sin z and cos z are unbounded.

One can use the previous identities to show

| sin z|2 = sin2 x + sinh2 y

and
| cos z|2 = cos2 x + sinh2 y .

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Exercise

Ex. Show that both sin z and cos z are unbounded.

One can use the previous identities to show

| sin z|2 = sin2 x + sinh2 y

and
| cos z|2 = cos2 x + sinh2 y .

Both are unbounded since the hyperbolic function sinh2 y is unbounded.

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